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anonymous

  • 5 years ago

Does the following series converge or diverge?

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  1. anonymous
    • 5 years ago
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    \[\sum_{n=2}^{infinity} 1/n \sqrt{\ln(n)}\]

  2. anonymous
    • 5 years ago
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    I'm wanting to say it converges to 1, but I'm not sure how to prove that. Real analysis wasn't my strong point.

  3. anonymous
    • 5 years ago
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    do you know what test to use?

  4. anonymous
    • 5 years ago
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    my original thought was ratio, but i dont know how to apply that here

  5. anonymous
    • 5 years ago
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    I'm wanting to say it has something to do with the Cauchy Criterion

  6. anonymous
    • 5 years ago
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    ...I have no idea what that means..haha

  7. anonymous
    • 5 years ago
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    I don't think we covered that

  8. anonymous
    • 5 years ago
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    Oh, the ratio test should work.

  9. anonymous
    • 5 years ago
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    really? Let me try that really quickly

  10. anonymous
    • 5 years ago
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    Well, that gave me infinity over infinity...maybe i'm doing it wrong?

  11. anonymous
    • 5 years ago
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    Looking up properties of the natural log right now, give me a sec.

  12. anonymous
    • 5 years ago
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    thanks so much!

  13. anonymous
    • 5 years ago
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    Integral test would be easier since i see ln(n) and 1/n there.

  14. anonymous
    • 5 years ago
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    oh that makes sense!

  15. anonymous
    • 5 years ago
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    Ok all i can think of is that \[n / (n+1) \le 1\] and \[\sqrt(\ln(n+1)) \le n\] and \[\sqrt(\ln(n)) \le n\] thus \[\left| n/(n+1) * \sqrt(\ln(n+1)) / \sqrt(\ln(n)) \right| \le \left| 1 * n/n \right| \le 1\]

  16. anonymous
    • 5 years ago
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    hmm. Ok, let me think about that for a min :)

  17. anonymous
    • 5 years ago
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    hmm actually that might not work cause the ration needs to be less than 1 for the ratio test to work

  18. anonymous
    • 5 years ago
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    when i worked it doing the integral test i got infinity meaning divergent does that seem right?

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